h3 - 40h2 + 375h - 1000 (h3 is h to the power of 3) - (h2 is h to the power of 2)
I know the general method - finding factors of 1000 and substituting them instead of h (factor theorem) - but the numbers are way too high. Is there any other method or do I just have to keep on substituting numbers?
(x2 + x - 12) / (x2+5x+6) < 0
(x+1)(x-2)(2x) >= 0 (2x is 2 to the power of x)
I know the general method - finding factors of 1000 and substituting them instead of h (factor theorem) - but the numbers are way too high. Is there any other method or do I just have to keep on substituting numbers?
(x2 + x - 12) / (x2+5x+6) < 0
(x+1)(x-2)(2x) >= 0 (2x is 2 to the power of x)
-
H^3-40h^2+375h-1000
Wow ,these are large numbers. First I would try factor by grouping.
H^2(h-40)+ 125(3h- 8)
H( h^2+375)-40(h^2+25)
Neither one worked.
So try synthetic division: factors of 1000?
+/- 1 do not work by observation., 2 and -2 are too small
5)1 -40 375 -1000
---------------------
....1 -35. 200 (0. ... Since the remainder is zero, (x-5) is a factor.
AND the other factor is (x^2-35x+200)
Once you get to second degree, you either factor or use quadratic formula...
It is tedious, but you can use your calculator in the real world. On tests they usually give smaller numbers:)
2) in solving rational inequalities, set up a line test using all of the zeroes of the numerators and denominators:
(x^2+x-12)/(x^2+5x+6)<0
(x+4)(x-3)/(x+2)(x+3)<0
.....(-4).....(-3)......(-2).....(3)..… This creates five regions to test. Try any number in each interval. You only care about the signs! You want negative results (<0)
f(-10)= (-)(-)/(-)(-)=(+)
f(-3.5)=(+)(-)/(-)(-)=(-)
f(-2.5)=(+)(-)/(-)(+)=(+)
f(0)=(+)(-)/(+)(+)=(-)
f(10)=(+)(+)/(+)(+)=(+)
So the solutions for negative results are in the second and fourth intervals:
(-4,-3) or (-2,3)
3) since 2^x is always positive, do not consider it for the problem.
......(-1).....(2)...... >=0
testing the three regions:
..(+).....(-).......(+)
X<=-1 or x>=2. Be sure to include the zeroes. ( because of the equal signs)
Hoping this helps!
Wow ,these are large numbers. First I would try factor by grouping.
H^2(h-40)+ 125(3h- 8)
H( h^2+375)-40(h^2+25)
Neither one worked.
So try synthetic division: factors of 1000?
+/- 1 do not work by observation., 2 and -2 are too small
5)1 -40 375 -1000
---------------------
....1 -35. 200 (0. ... Since the remainder is zero, (x-5) is a factor.
AND the other factor is (x^2-35x+200)
Once you get to second degree, you either factor or use quadratic formula...
It is tedious, but you can use your calculator in the real world. On tests they usually give smaller numbers:)
2) in solving rational inequalities, set up a line test using all of the zeroes of the numerators and denominators:
(x^2+x-12)/(x^2+5x+6)<0
(x+4)(x-3)/(x+2)(x+3)<0
.....(-4).....(-3)......(-2).....(3)..… This creates five regions to test. Try any number in each interval. You only care about the signs! You want negative results (<0)
f(-10)= (-)(-)/(-)(-)=(+)
f(-3.5)=(+)(-)/(-)(-)=(-)
f(-2.5)=(+)(-)/(-)(+)=(+)
f(0)=(+)(-)/(+)(+)=(-)
f(10)=(+)(+)/(+)(+)=(+)
So the solutions for negative results are in the second and fourth intervals:
(-4,-3) or (-2,3)
3) since 2^x is always positive, do not consider it for the problem.
......(-1).....(2)...... >=0
testing the three regions:
..(+).....(-).......(+)
X<=-1 or x>=2. Be sure to include the zeroes. ( because of the equal signs)
Hoping this helps!